0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
A catalyst
A catalyst can be in many forms
Answer:
heat flows from the object that has more thermal more energy to the object with less thermal energy
The main points of Dalton's atomic theory are: Everything is composed of atoms, which are the indivisible building blocks of matter and cannot be destroyed. All atoms of an element are identical. The atoms of different elements vary in size and mass.
Answer:
- <em>Oxidation half-reaction</em>:
Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻
- <em>Reduction half-reaction</em>:
Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)
Explanation:
The reaction that takes place is:
- Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq)
The <em>oxidation half-reaction</em> is:
- Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻
It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.
The <em>reduction half-reaction</em> is:
- Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)
It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.
